Mod 4

Let $R_1, R_2, \dots, R_5$ be the products of the numbers in the first row, second row, etc. respectively. Define $C_1, C_2, \dots, C_5$ similarly with the products of the numbers in the columns.

We have $\prod_{i = 1}^5 R_i = \prod_{i = 1}^5 C_i,$ since every integer in the square is represented in both products. However,

$\prod_{i = 1}^5 R_i \equiv 1^5 \equiv 1 \! \! \! \! \pmod{4},$

and

$\prod_{i = 1}^5 C_i \equiv 3^5 \equiv 3 \! \! \! \! \pmod{4}.$

No positive integer can simultaneous have both a remainder of 1 and 3 when divided by 4. Therefore, we conclude that no , it is not possible to fill in each cell of the 5 x 5 grid such that the products of each row are 1 modulo 4, and the products of each column are 3 modulo 4.

The core of this solution is the same as the one already posted, but I share it as it is a way to transform this problem into another problem that is much more intuitive:

What needs to be proven can actually be deduced from this intuitive fact: There can't exist a $5 \times 5$ matrix where the sum of the elements of each column is odd and the sum of the elements of each row is even. To show this first suppose that a matrix $M$ with the properties given in the problem exists. It can be easily shown that all the elements of this matrix must be either $1 \mod 4$ or $3 \mod 4$ . (Assume there is an element congruent to 0, then the product of the elements in the row where that element is will have remainder 0. If there was an element congruent to 2, the product of the elements in that row will have remainder 2 or 0).

So now take this matrix $M$ to the matrix $F(M)$ where $F$ changes every number congruent to 1 with $1$ and every number congruent to 3 to $-1$ . (Remember $-1 \equiv 3 \mod 4$ . This new matrix has the properties:

1: The product of each row is 1

2: The product of each column is -1

Now, as all the elements are either $1's$ or $(-1)'s$ , all elements can be written in the form $(-1)^a$ . Next take the matrix $F(M)$ to $G(F(M))$ where $G$ maps every element of the form $(-1)^a$ to $a$ , $a \in \{0,1 \}$ . (In other words, G(1) = 0, G(-1) = 1). This new matrix has the following properties:

1: The sum of each row is even

2: The sum of each column is odd

But no such matrix can exist, so we have a contradiction. Notice that the previous construction turned a problem of congruences $\mod 4$ into a problem of congruences $\mod 2$ .